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2 years ago

I need help with this please!!

Mathematics

2 answers:

nekit [7.7K]2 years ago

8 0

Answer:

A & C

Step-by-step explanation:

A) 8(3j-2)

= 24j-16

B) 6(4j-2j)

= 24j -12 j (not equal)

C) 4 (6j-4)

24j -16

D) 2(12-8)

24-16 (not equal)

Therefore only A and C are equivalent to 24j-16

Answered by Gauthmath

topjm [15]2 years ago

3 0

Answer:

8(3j -2)

4(6j -4)

Step-by-step explanation:

24j - 16

Rewriting

8*3*j - 8*2

Factor out the greatest common factor

8(3j -2)

We can also factor out 4

4*6j - 4*4

4(6j -4)

You might be interested in

It is possible to get 2 solutions when the system of equations is: (check all that apply)

SVETLANKA909090 [29]

Answer:

A system of the equation of a circle and a linear equation

A system of the equation of a parabola and a linear equation

Step-by-step explanation:

Let us verify our answer

A system of the equation of a circle and a linear equation

Let an equation of a circle as $x^2+ y^2 = 1$ ..........(1)

Let a liner equation Y = x ............(2)

substitute (2) in (1)

$x^2 + x^2 = 1\\2x^2 = 1\\$

$x^2 = \frac{1}{\sqrt{2} } \\x = +\frac{1}{\sqrt{2} } , -\frac{1}{\sqrt{2} }$ so Y = $+\frac{1}{\sqrt{2} } , -\frac{1}{\sqrt{2} }$

so the two solution are ( $(\frac{1}{\sqrt{2} } ,\frac{1}{\sqrt{2} }) (-\frac{1}{\sqrt{2} }, -\frac{1}{\sqrt{2} }$ )

A system of the equation of a parabola and a linear equation

Let equation of Parabola be $y^2 = x$

and linear equation y = x

substitute

$x^2 = x\\x^2 - x= 0\\x(x-1) = \\x = 0 , 1$

Y = 0,1

so the two solutions will be (0,0) and (1,1)

3 0

3 years ago

The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a

Leya [2.2K]

Answer:

a) 0.25

b) 52.76% probability that a person waits for less than 3 minutes

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \lambda e^{-\lambda x}$

In which $\lambda = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

In this question:

$m = 4$

a. Find the value of λ.

$\lambda = \frac{1}{m} = \frac{1}{4} = 0.25$

b. What is the probability that a person waits for less than 3 minutes?

$P(X \leq 3) = 1 - e^{-0.25*3} = 0.5276$

52.76% probability that a person waits for less than 3 minutes

3 0

3 years ago

Hellppp meeeeeee PLEASEEEEE PLEASEEEEE hellppp meeeeeee

Dimas [21]

Answer:

5 min

Step-by-step explanation:

I think its 17- 2ft (which was already dug)

then 15÷3 which is 5

4 0

3 years ago

Read 2 more answers

Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2

Finger [1]

Answer:

$\frac{y}{x^2}=\sin x+\pi$

Step-by-step explanation:

Consider linear differential equation $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$

It's solution is of form $y\,I.F=\int I.F\,q(x)\,dx$ where I.F is integrating factor given by $I.F=e^{\int p(x)\,dx}$ .

Given: $\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x$

We can write this equation as $\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x$

On comparing this equation with $\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)$ , we get $p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x$

I.F = $e^{\int p(x)\,dx}=e^{\int \frac{-2}{x}\,dx}=e^{-2\ln x}=e^{\ln x^{-2}}=\frac{1}{x^2}$ { formula used: $\ln a^b=b\ln a$ }

we get solution as follows:

$\frac{y}{x^2}=\int \frac{1}{x^2}x^2\cos x\,dx\\\frac{y}{x^2}=\int \cos x\,dx\\\\\frac{y}{x^2}=\sin x+C$

{ formula used: $\int \cos x\,dx=\sin x$ }

Applying condition: $y(\pi)=\pi^2$

$\frac{y}{x^2}=\sin x+C\\\frac{\pi^2}{\pi}=\sin\pi+C\\\pi=C$

So, we get solution as :

$\frac{y}{x^2}=\sin x+\pi$

4 0

3 years ago

explain the benefits of each of the three forms of quadratic equations, standard form, vertex form, and factored form. What do t

mamaluj [8]

Answer:

Summary:

Standard form allow us to quickly find the y-intercept.

Vertex form allow us to quickly locate the vertex.

And factored form allows us to quickly determine the roots/zeros.

Step-by-step explanation:

The three forms of quadratics are the standard form, vertex form, and the factored form. Each of them reveals a specific part about the quadratic.

Standard Form:

The standard form of a quadratic is:

$ax^2+bx+c$

There are only two details that can be conveyed by a quadratic in standard form immediately: (1) the leading coefficient a, and (2) the y-intercept.

The leading coefficient a will tell us if the parabola curves upwards or downwards.

And the constant c will give us the y-intercept.

Vertex Form:

The vertex form of a quadratic is:

$a(x-h)^2+k$

There are also two details that can be conveyed by a quadratic in vertex form: (1) the vertex, and (2) the leading coefficient.

The leading coefficient is given by a. Again, this tells us the orientation of the parabola.

And the vertex is given by (h, k).

Hence, in my opinion, vertex form is the best form of a quadratic since it immediately reveals the vertex, the most important aspect of a quadratic.

Factored Form:

The factored form of a quadratic is:

$a(x-p)(x-q)$

Where p and q are the zeros/roots/solutions of the quadratic.

Again, factored form gives us two details about the quadratic: (1) the leading coefficient, and (2) the zeros.

The zeros tells us when the parabola crosses the x-axis, which can assist in graphing.

Summary:

Therefore, each form of a quadratic equation has its own benefits.

Standard form allow us to find the y-intercept.

Vertex form allow us to quickly locate the vertex.

And factored form allows us to quickly determine the roots/zeros.

Hence, depending on the question, each form can be useful in its own way.

4 0

3 years ago